c - 为什么一些库例程同时实现为宏？为什么 "va_arg"宏被声明为函数(没有 "#define")？

Question

我很难用语言清楚地表达出来。所以让我把它分成几部分。上下文来自 Mike Banahan 的 C 书籍(下面每个部分都提供了链接)。以下是我的问题，以粗体显示:

• 为什么有些库函数也同时实现为宏？需要什么？这是我从书中读到的内容(Section 9.1.1) :

A last general point is that many of the library routines may be implemented as macros, provided that there will be no problems to do with side-effects (as Chapter 7 describes). The Standard guarantees that, if a function is normally implemented as a macro, there will also be a true function provided to do the same job. To use the real function, either undefine the macro name with #undef, or enclose its name in parentheses, which ensures that it won't be treated as a macro:

• va_start，正如我们所知，是函数还是宏？书中的以下文字是困惑的根源，因为它在同一呼吸中，在相邻的行中暗示了两者! (Section 9.9)

Before any attempt can be made to access a variable argument list, va_start must be called. It is defined as

 #include <stdarg.h>
void va_start(va_list ap, parmN); 

The va_start macro initializes ap for subsequent use by the functions va_arg and va_end.

• 最后是最令人困惑的部分。在下面的行中，清楚地写着va_arg是一个宏，并继续展示它是如何实现的。但是，如何在没有 #define 关键字的情况下实现宏，并且也像函数一样具有返回类型('type')呢？ (Section 9.9)

Once initialized, the arguments supplied can be accessed sequentially by means of the va_arg macro. This is peculiar because the type returned is determined by an argument to the macro. Note that this is impossible to implement as a true function, only as a macro. It is defined as

#include <stdarg.h>
type va_arg(va_list ap, type);

我们将非常感谢您的回答。谢谢。

Answer 1

这是两个不同的问题。

首先，va_start、va_arg和va_end保证是宏。 va_arg 不能是函数，因为它的第二个参数是类型而不是值。

至于为什么有些函数也是宏:因为您希望内联调用，使其更快，但您可能还想获取其地址并将其放入函数指针中:

int (*function_pointer)(int, FILE *) = &(putc);